Here is a solution taken from pages 265 and 266 of a Chinese document. The solution is not mine; I just give a translation here.
There are some mistakes in the original solution (in Chinese); I corrected them.
Suppose that $n > 1$.
Adding $-1$ times row $n-1$ to row $n$,
adding $-1$ times row $n-2$ to row $n-1$,
...,
adding $-1$ times row $2$ to row $1$,
we yield
$$
S_2 =
\begin{bmatrix}
0 & 1+2 \mathrm{i} & 1+3 \mathrm{i} & \cdots & \cdots & 1+(n-1) \mathrm{i} & 1+n \mathrm{i} \\
1+2 \mathrm{i} & -(1+2 \mathrm{i}) & 1 & \cdots & \cdots & 1 & 1 \\
\mathrm{i} & 2+3 \mathrm{i} & -(2+3 \mathrm{i}) & \ddots & & \vdots & \vdots \\
\vdots & \vdots & \ddots & \ddots & \ddots & \vdots & \vdots \\
\vdots & \vdots & & \ddots & \ddots & 1 & 1 \\
\mathrm{i} & \mathrm{i} & \cdots & \cdots & (n-2)+(n-1) \mathrm{i} & -((n-2)+(n-1) \mathrm{i}) & 1 \\
\mathrm{i} & \mathrm{i} & \cdots & \cdots & \mathrm{i} & (n-1)+n \mathrm{i} & -((n-1)+n \mathrm{i}) \\
\end{bmatrix}.
$$
Note that $\det {(S_2)} = \det {(S)}$.
Adding $-1$ times column $n-1$ to column $n$,
adding $-1$ times column $n-2$ to column $n-1$,
...,
adding $-1$ times column $2$ to column $1$,
we yield
$$
S_3 =
\begin{bmatrix}
0 & -\alpha_{1} & \mathrm{i} & \cdots & \cdots & \mathrm{i} & \mathrm{i} \\
-\alpha_{1} & 2 \alpha_{1} & 2 \alpha & 0 & \cdots & 0 & 0 \\
\mathrm{i} & 2 \alpha & 2 \alpha_{2} & \ddots & \ddots & \vdots & \vdots \\
\vdots & 0 & \ddots & \ddots & \ddots & 0 & 0 \\
\vdots & \vdots & \ddots & \ddots & \ddots & (n-2) \alpha & 0 \\
\mathrm{i} & 0 & \cdots & 0 & (n-2) \alpha & 2 \alpha_{n-2} & (n-1) \alpha \\
\mathrm{i} & 0 & \cdots & 0 & 0 & (n-1) \alpha & 2 \alpha_{n-1}
\end{bmatrix},
$$
in which $\alpha = 1 + \mathrm{i}$, $\alpha_k = -(k + (k + 1)\mathrm{i})$ for $k = 1$, $2$, $\dots$, $n-1$.
Note that $\det {(S_3)} = \det {(S_2)} = \det {(S)}$.
Multiplying row $1$ by $\alpha$,
multiplying column $1$ by $\alpha$,
we yield
$$
S_4 =
\begin{bmatrix}
0 & -\alpha_{1} \alpha & \mathrm{i} \alpha & \cdots & \cdots & \mathrm{i} \alpha & \mathrm{i} \alpha \\
-\alpha_{1} \alpha & 2 \alpha_{1} & 2 \alpha & 0 & \cdots & 0 & 0 \\
\mathrm{i} \alpha & 2 \alpha & 2 \alpha_{2} & \ddots & \ddots & \vdots & \vdots \\
\vdots & 0 & \ddots & \ddots & \ddots & 0 & 0 \\
\vdots & \vdots & \ddots & \ddots & \ddots & (n-2) \alpha & 0 \\
\mathrm{i} \alpha & 0 & \cdots & 0 & (n-2) \alpha & 2 \alpha_{n-2} & (n-1) \alpha \\
\mathrm{i} \alpha & 0 & \cdots & 0 & 0 & (n-1) \alpha & 2 \alpha_{n-1}
\end{bmatrix}.
$$
Note that $\det {(S_4)} = \frac{1}{\alpha^2} \det {(S_3)} = \frac{1}{\alpha^2} \det {(S)}$.
Add all rows except the first to row $1$.
Add all columns except the first to column $1$.
We yield
$$
S_5 =
\begin{bmatrix}
2(-n+\mathrm{i}) & -1+\mathrm{i} & 0 & \cdots & \cdots & 0 & -n \alpha \\
-1+\mathrm{i} & 2 \alpha_{1} & 2 \alpha & 0 & \cdots & 0 & 0 \\
0 & 2 \alpha & 2 \alpha_{2} & \ddots & \ddots & \vdots & \vdots \\
\vdots & 0 & \ddots & \ddots & \ddots & 0 & 0 \\
\vdots & \vdots & \ddots & \ddots & \ddots & (n-2) \alpha & 0 \\
0 & 0 & \cdots & 0 & (n-2) \alpha & 2 \alpha_{n-2} & (n-1) \alpha \\
-n \alpha & 0 & \cdots & 0 & 0 & (n-1) \alpha & 2 \alpha_{n-1}
\end{bmatrix}.
$$
Note that $\det {(S_5)} = \det {(S_4)} = \frac{1}{\alpha^2} \det {(S)}$.
We will show that the determinant of $S_5$ is nonzero.
Since
$$
\begin{aligned}
& |2(-n + \mathrm{i})|^2 - (|{-1 + \mathrm{i}}| + |{-n\alpha}|)^2 \\
= {} & 4(n^2 + 1) - (\sqrt{2} + \sqrt{2}n)^2 \\
= {} & 2(n - 1)^2 > 0,
\end{aligned}
$$
we have
$$
|2(-n + \mathrm{i})| > |{-1 + \mathrm{i}}| + |{-n\alpha}|.
$$
Since
$$
\begin{aligned}
& |2\alpha_1|^2 - (|{-1 + \mathrm{i}}| + |{2\alpha}|)^2 \\
= {} & 20 - 18 > 0,
\end{aligned}
$$
we have
$$
|2\alpha_1| > |{-1 + \mathrm{i}}| + |{2\alpha}|.
$$
Since
$$
\begin{aligned}
& |2\alpha_k|^2 - (|k\alpha| + |{(k+1)\alpha}|)^2 \\
= {} & 4(k^2 + (k+1)^2) - (\sqrt{2}k + \sqrt{2}(k+1))^2 \\
= {} & 2 > 0,
\end{aligned}
$$
we have
$$
|2\alpha_k| > |{k\alpha}| + |{(k+1)\alpha}|
$$
for $k = 2$, $3$, $\dots$, $n-1$.
According to Levy-Desplanques theorem, the determinant of $S_5$ is nonzero.
Hence that of $S$ is nonzero.
